I'm trying to compare the week number of a started task with the week before the current one.
and to_char(t.plan_start_dttm, 'YYYY-IW') = to_char(sysdate-1, 'YYYY-IW')
So yesterday, it was working fine, I got all records of last week. But today, I get the ones of this week.
So apparently, my "sysdate -1 is not the right way to do it.
Thank you
yesterday, it was working fine, I got all records of last week.
Yesterday was Monday of this week; so subtracting one day would be Sunday of last week and truncating to the start of the week would be Monday of last week.
Today is Tuesday; so subtracting one day would be Monday of this week and then truncating to the start of the week would still be this week.
Instead, subtract an entire week's worth of days (7):
and t.plan_start_dttm >= TRUNC( sysdate - 7, 'IW')
and t_plan_start_dttm < TRUNC( sysdate, 'IW' )
and if you compare on a range then Oracle can use an index on the plan_start_dttm column; whereas if you use TO_CHAR or TRUNC on the column then the column index cannot be used and you would need a separate function-based index.
Subtract 7 days instead of 1. Or 1 week. I also prefer trunc():
trunc(t.plan_start_dttm, 'IW') = trunc(sysdate - interval '7' day, 'IW')
There is no need to convert to strings for this purpose.
Related
I'm sorry if the title is confusing.
Basically I am creating a dashboard using Oracle Apex 18.x and in one my card, I am comparing the sales between today vs yesterday, this week vs last week, this month vs last month and this year vs last year.
I have no problem with the today vs yesterday. My issue is with the week, month and year because I want to compare only up to the current day, not the whole last week.
For example, if today is the 4th day of the current week, then I'd have to compare it from last week up to its 4th day only as well.
For the month, if let's say today is 25-Mar-2018 then I should only compare it to the month of Feb from 1-25 only. Same with the year.
It wouldn't be a problem if the requirement is to get the full last week, full last month and full last year.
For week i would do it like:
select
TRUNC(sysdate, 'DAY') start_of_the_week,
(sysdate) day_of_current_week,
(TRUNC(sysdate, 'DAY') - 7 ) start_of_last_week
(sysdate - 7) day_of_last_week,
from dual;
For month it would be:
select
(sysdate) day_of_current_month,
TRUNC(sysdate, 'mm') month_start_date,
ADD_MONTHS(sysdate,-1) day_of_last_month,
ADD_MONTHS(TRUNC(sysdate, 'mm'),-1) start_of_last_month
from dual;
I have spent days trying to figure this out to no avail, so hopefully someone can help me. I have a queried date set which contains several fields including a column of dates. What I want to do is create a new field in my query that tells what the Monday and Friday is for the week of that row's particular date.
So for example; if the date in one of my rows is "1/16/18",
the new field should indicate "1/15/18 - 1/19/18".
So basically I need to be able to extract the Monday date (1/15/18) and the Friday date (1/19/18) of the week of 1/16/18 and then concatenate the two with a dash ( - ) in between. I need to do this for every row.
How on earth do I do this? I've been struggling just to figure out how to find the Monday or Friday of the given date...
Assuming that your column is of type date, you can use trunc to get the first day of the week (monday) and then add 4 days to get the friday.
For example:
with yourTable(d) as (select sysdate from dual)
select trunc(d, 'iw'), trunc(d, 'iw') + 4
from yourTable
To format the date as a string in the needed format, you can use to_char; for example:
with yourTable(d) as (select sysdate from dual)
select to_char(trunc(d, 'iw'), 'dd/mm/yy') ||'-'|| to_char(trunc(d, 'iw') + 4, 'dd/mm/yy')
from yourTable
gives
15/01/2018-19/01/18
There may be a simpler, canonical Oracle method to this but you can still reduce it to a simple calculation on your own either way. I'm going to assume you're dealing with only dates falling Monday through Friday. If you do need to deal with weekend dates then you might have to be more explicit about which logical week they should be attached to.
<date> - (to_char(<date>, 'D') - 2) -- Monday
<date> + (6 - to_char(<date>, 'D')) -- Friday
In principle all you need to do is add/subtract the appropriate number of days based on the current day of week (from 1 - 7). There are some implicit casts going on in there and it would probably be wise to handle those better. You might also want to check into NLS settings to make sure you can rely on to_char() using Sunday as the first day of week.
https://docs.oracle.com/cd/B19306_01/server.102/b14200/sql_elements004.htm
You can also use the NEXT_DAY function, as in:
SELECT TRUNC(NEXT_DAY(SYSDATE, 'MON')) - INTERVAL '7' DAY AS PREV_MONDAY,
TRUNC(NEXT_DAY(SYSDATE, 'FRI')) AS NEXT_FRIDAY
FROM DUAL;
Note that using the above, on weekends the Monday will be the Monday preceding the current date, and the Friday will be the Friday following the current date, i.e. there will be 11 days between the two days.
You can also use
SELECT TRUNC(NEXT_DAY(SYSDATE, 'MON')) - INTERVAL '7' DAY AS PREV_MONDAY,
TRUNC(NEXT_DAY(SYSDATE, 'MON')) - INTERVAL '3' DAY AS NEXT_FRIDAY
FROM DUAL;
in which case the Monday and Friday will always be from the same week, but if SYSDATE is on a weekend the Monday and Friday returned will be from the PREVIOUS week.
I have a date in sql which will always fall on a Monday and I'm subtracting 13 weeks to get a weekly report. I am trying to get the same 13 week report but for last year's figures as well.
At the moment, I'm using the following:
calendar_date >= TRUNC(sysdate) - 91
which is working fine.
I need the same for last year.
However, when I split this into calendar weeks, there will also be a partially complete week as it will include 1 or 2 days from the previous week. I need only whole weeks.
e.g. the dates that will be returned for last year will be 14-Feb-2015 to 16-May-2015. I need it to start on the Monday and be 16-Feb-2015. This will change each week as I am only interested in complete weeks...
I would do this:
Get the date by substracting 91 days as you're already doing.
Get the number of the day of the week with TO_CHAR(DATE,'d')
Add the number of days until the next monday to the date.
Something like this:
SELECT TO_DATE(TO_DATE('16/05/2015','DD/MM/YYYY'),'DD/MM/YYYY')-91 + MOD(7 - TO_NUMBER(TO_CHAR(TO_DATE(TO_DATE('16/05/2015','DD/MM/YYYY'),'DD/MM/RRRR')-91,'d'))+1,7) d
FROM dual
next_day - returns date of first weekday named by char.
with dates as (select to_date('16/05/2015','DD/MM/YYYY') d from dual)
select
trunc(next_day( trunc(d-91) - interval '1' second,'MONDAY'))
from dates;
I want to get next monday from calculated date. In situation when calculated date is monday i have to move back to previous week ( -1 second).
Need help figuring out how to determine if the date is the same 'day' as today in teradata. IE, today 12/1/15 Tuesday, same day last year was actually 12/2/2014 Tuesday.
I tried using current_date - INTERVAL'1'Year but it returns 12/1/2014.
You can do this with a bit of math if you can convert your current date's "Day of the week" to a number, and the previous year's "Day of the week" to a number.
In order to do this in Teradata your best bet is to utilize the sys_calendar.calendar table. Specifically the day_of_week column. Although there are other ways to do it.
Furthermore, instead of using CURRENT_DATE - INTERVAL '1' YEAR, it's a good idea to use ADD_MONTHS(CURRENT_DATE, -12) since INTERVAL arithmetic will fail on 2012-02-29 and other Feb 29th leap year dates.
So, putting it together you get what you need with:
SELECT
ADD_MONTHS(CURRENT_DATE, -12)
+
(
(SELECT day_of_week FROM sys_calendar.calendar WHERE calendar_date = CURRENT_DATE)
-
(SELECT day_of_week FROM sys_calendar.calendar WHERE calendar_date = ADD_MONTHS(CURRENT_DATE, -12))
)
This is basically saying: Take the current dates day of week number (3) and subtract from it last years day of week number (2) to get 1. Add that to last year's date and you'll have the same day of the week as current date.
I tested this for all dates between 01/01/2010 and CURRENT_DATE and it worked as expected.
Why don't you simply subtract 52 weeks?
current_date - 364
The SQL below will get you to the abbreviated name for the day of week, it's cumbersome but it works across versions of Teradata.
SELECT CAST(CAST(ADD_MONTHS(CURRENT_DATE, -12) AS DATE FORMAT 'E3') AS CHAR(3)) AS LY_DayOfWeek
, CAST(CAST(CURRENT_DATE) AS DATE FORMAT 'E3') AS CHAR(3)) AS CY_DayOfWeek
Dates are internally represented at integers in Teradata as (Year-1900) * 100000 + (MONTH * 100) + DAY. You may be able to do some creative arithmetic to figure out that 12/1/2015 Tuesday was 12/2/2014 Tuesday last year.
I'm keep track of recurring weekly events in a table using just a DATETIME. I only care about the TIME and the day of the week it falls on.
I need to be able to convert the set DATETIME into the current or upcoming future one.
IE How can I convert a date stored as 2013-02-22 12:00:00 using the current date to the next occurrence? Ie this next Friday at 12:00:00 or 2013-03-01 12:00:00 so that I can then order events by date?
Or I could store the TIME and day of the week separately as a number 0-6.
UPDATE:
From Erwin I got something like:
Event.order("date_trunc('week', now()::timestamp) + (start_at - date_trunc('week', start_at))")
Which seems order them except that the first dates I get are Monday skipping over events I know exist for Sunday which it puts as last.
Your best choice is to store a timestamp or timestamptz (timestamop with time zone). If you have or ever will have to deal with more than one time zone, make that timestamptz and define whether you want to operate with local time or UTC or whatever. More details in this related answer:
Ignoring timezones altogether in Rails and PostgreSQL
Demo how to transpose a timestamp into the current week efficiently (same day of week and time). Assuming timestamp here:
SELECT date_trunc('week', now()::timestamp) + (t - date_trunc('week', t))
FROM (SELECT '2013-02-15 12:00:00'::timestamp AS t) x;
The trick is to compute the interval between the start of the corresponding week and the given timestamp and add that to the start of the current week with the help of date_trunc().
The ISO week starts with Monday, putting Sunday last.
Or, to just add a week to a given timestamp:
SELECT t + interval '1 week';
If You just want to ORDER BY, you only need the interval:
ORDER BY (t - date_trunc('week', t))
If you want to put Sunday first (shifting days):
ORDER BY ((t + interval '1d') - date_trunc('week', (t + interval '1d'))
Or simpler:
ORDER BY EXTRACT(dow FROM t), t::time
Quoting the manual on EXTRACT():
dow
The day of the week as Sunday(0) to Saturday(6)
isodow
The day of the week as Monday(1) to Sunday(7)
Answer to question in comment
I'm only interested in ordering them relative to the current date. Ie
if it's tuesday, I want tuesday first, monday last.
Wrapping at midnight of "today":
ORDER BY (EXTRACT(dow FROM t)::int + 7 - EXTRACT(dow FROM now())::int) % 7
,t::time
Using the modulo operator % to shift the day according to "today".
Using dowinstead of isodow, because starting with 0 makes % simpler.
Keep using the datetime. It's simple and gives you flexibility. You can use the extract function to get your time of day and day of week results. This page will help you. http://www.postgresql.org/docs/9.3/static/functions-datetime.html